Measure a curve with a chain of tiny straight segments, each of length $\sqrt{1+(y')^2}\,dx$ — then spin those segments into bands to find the area of a surface.
How long is a hanging power cable between two pylons? How much sheet metal is needed to press a satellite dish, or how much glaze covers a vase turned on a potter's wheel? These questions ask for the length of a curve and the area of a curved surface. Neither can be read off from a formula for straight lines and flat pieces — but both can be built from them. A curve looks almost straight when we zoom in far enough, so we approximate it by a chain of short line segments, add their lengths, and let the segments shrink. The sum becomes an integral.
The same idea, spun around an axis, gives surface area: each short segment sweeps out a thin band shaped like a lampshade (a frustum of a cone), whose area we know exactly. In this section we turn both constructions into formulas, meet the hyperbolic cosine — the exact shape of a hanging chain — and see how numerical integration rescues us when, as often happens, the arc-length integral has no elementary antiderivative.
Arc Length of a Graph
Let $f$ have a continuous derivative on $[a,b]$. Partition $[a,b]$ into $n$ pieces of width $\Delta x$ and join the points $P_i=\big(x_i,f(x_i)\big)$ by straight segments. The $i$-th segment has length
$$|P_{i-1}P_i|=\sqrt{(\Delta x)^2+(\Delta y_i)^2}=\sqrt{1+\left(\frac{\Delta y_i}{\Delta x}\right)^2}\,\Delta x .$$
By the Mean Value Theorem there is a point $x_i^*$ in $[x_{i-1},x_i]$ with $\dfrac{\Delta y_i}{\Delta x}=f'(x_i^*)$. So the polygon has length
$$\sum_{i=1}^{n}\sqrt{1+\big(f'(x_i^*)\big)^2}\,\Delta x ,$$
a Riemann sum for the continuous function $\sqrt{1+\big(f'(x)\big)^2}$. Its limit is the length of the curve.
📘 Definition 6.5 — Smooth curve and arc length
A function $f$ is smooth on $[a,b]$ if $f'$ exists and is continuous on $[a,b]$. The arc length of the graph of a smooth function from $x=a$ to $x=b$ is the limit of the lengths of inscribed polygons as the largest segment shrinks to zero.
📐 Theorem 6.7 — Arc-length formulas
If $f$ is smooth on $[a,b]$, the length of the curve $y=f(x)$, $a\le x\le b$, is
$$L=\int_a^b\sqrt{1+\big(f'(x)\big)^2}\,dx=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx .$$
If instead the curve is $x=g(y)$, $c\le y\le d$, with $g$ smooth on $[c,d]$, then
$$L=\int_c^d\sqrt{1+\big(g'(y)\big)^2}\,dy=\int_c^d\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy .$$
Both come from the "Pythagoras for differentials" $ds^2=dx^2+dy^2$.
✏️ Example 1 — A semicubical parabola
Find the length of the curve $y=x^{3/2}$ from $x=0$ to $x=4$.
💡 Show solution▶
Derivative $f'(x)=\tfrac32x^{1/2}$, so $1+\big(f'(x)\big)^2=1+\tfrac94x$. ($f'$ is continuous on $[0,4]$.)
Integrate With $w=1+\tfrac94x$, $dw=\tfrac94dx$ ($x=0\Rightarrow w=1$, $x=4\Rightarrow w=10$): $$L=\int_0^4\sqrt{1+\tfrac94x}\,dx=\frac49\int_1^{10}w^{1/2}\,dw=\frac49\cdot\frac23\Big[w^{3/2}\Big]_1^{10}=\frac{8}{27}\left(10\sqrt{10}-1\right)\approx9.0734 .$$
Check The straight line from $(0,0)$ to $(4,8)$ has length $\sqrt{16+64}\approx8.944$. The curve must be a little longer, and $9.073\gt8.944$ ✓. Interactive 1 shows the polygon lengths climbing to $9.0734$.
Interactive 1 — Polygons Converging to the Arc Length
The red polygon joins $n+1$ points on the curve. One segment is highlighted with its legs $\Delta x$ and $\Delta y$, so its length is $\sqrt{\Delta x^2+\Delta y^2}$. Press ▶ to double the number of segments: the polygon length always increases towards the exact arc length $L$, and never passes it.
2
segments $n$—
polygon length $L_n$—
exact $L$—
$L-L_n$—
✏️ Example 2 — The perfect-square trick
Find the length of $y=\dfrac{x^3}{6}+\dfrac{1}{2x}$ for $1\le x\le2$.
Spot the square Adding $1$ flips the sign of the middle term: $$1+\big(f'(x)\big)^2=\frac{x^4}{4}+\frac12+\frac{1}{4x^4}=\left(\frac{x^2}{2}+\frac{1}{2x^2}\right)^2 .$$ Since $\tfrac{x^2}{2}+\tfrac{1}{2x^2}\gt0$, the square root is just $\tfrac{x^2}{2}+\tfrac{1}{2x^2}$.
Why it works Here $f'=u-v$ with $u=\tfrac{x^2}{2}$, $v=\tfrac{1}{2x^2}$ and $uv=\tfrac14$. Whenever $uv=\tfrac14$, $$1+(u-v)^2=u^2-2uv+v^2+4uv=(u+v)^2 ,$$ so the square root disappears. Many arc-length exercises are built this way.
⚠️ Common mistakes with arc length
1. "Simplifying" the square root. $\sqrt{1+(f')^2}$ is not $1+f'$. The root only disappears when $1+(f')^2$ is a perfect square.
2. Forgetting to square. The integrand is $\sqrt{1+(f')^2}$, not $\sqrt{1+f'}$.
3. Using $f$ instead of $f'$. Arc length depends on the slope, not the height: $y=x^2$ and $y=x^2+100$ have the same length on any interval.
4. Expecting a nice antiderivative. Most arc-length integrals — even for $y=\sin x$ or $y=x^3$ — cannot be done in closed form. Then a numerical method (Section 5.4) is the right tool, not a mistake.
The Arc-Length Function and the Catenary
Instead of a fixed upper limit, let the end point vary. The resulting function measures distance along the curve, like the kilometre markers along a winding road.
📘 Definition 6.6 — Arc-length function
If $f$ is smooth on $[a,b]$, the arc-length function of $y=f(x)$ starting at $x=a$ is
$$s(x)=\int_a^x\sqrt{1+\big(f'(t)\big)^2}\,dt,\qquad a\le x\le b .$$
By the Fundamental Theorem of Calculus, $\dfrac{ds}{dx}=\sqrt{1+\big(f'(x)\big)^2}\ge1$, so $s$ is increasing, and in differential form $ds=\sqrt{1+\left(\tfrac{dy}{dx}\right)^2}\,dx$, i.e. $(ds)^2=(dx)^2+(dy)^2$.
A flexible chain or cable hanging under its own weight does not form a parabola, as Galileo guessed, but a curve called a catenary (from the Latin catena, chain). Its equation uses a new function.
📘 Definition 6.7 — Hyperbolic cosine and sine
For all real $x$,
$$\cosh x=\frac{e^x+e^{-x}}{2},\qquad\sinh x=\frac{e^x-e^{-x}}{2}.$$
They satisfy $\dfrac{d}{dx}\cosh x=\sinh x$, $\dfrac{d}{dx}\sinh x=\cosh x$ and the identity $\cosh^2x-\sinh^2x=1$ (expand both squares to check). Note $\cosh x\ge1$ and $\cosh0=1$. The graph of $y=a\cosh(x/a)$, $a\gt0$, is a catenary.
✏️ Example 3 — Length of a catenary
(a) Find the arc-length function of $y=\cosh x$ starting at $x=0$. (b) Find the length of $y=\cosh x$ for $-1\le x\le1$.
💡 Show solution▶
Simplify $f'(x)=\sinh x$ and, by the identity, $1+\sinh^2x=\cosh^2x$. Since $\cosh x\gt0$, $\sqrt{1+\sinh^2x}=\cosh x$.
(a) $$s(x)=\int_0^x\cosh t\,dt=\sinh x .$$ The distance along the chain from its lowest point equals $\sinh x$ — the slope there. This neat fact is exactly why a hanging chain takes this shape.
(b) By symmetry, $$L=\int_{-1}^1\cosh x\,dx=\Big[\sinh x\Big]_{-1}^1=2\sinh1=e-\frac1e\approx2.3504 .$$
Check The chord from $(-1,\cosh1)$ to $(1,\cosh1)$ has length $2$; the sagging curve is longer, $2.3504\gt2$ ✓.
✏️ Example 4 — When to integrate in $y$
Find the length of $y=x^{2/3}$ from $(0,0)$ to $(8,4)$.
💡 Show solution▶
Problem in $x$ $f'(x)=\tfrac23x^{-1/3}$ is unbounded as $x\to0^+$, so $f$ is not smooth on $[0,8]$ and Theorem 6.7 (in $x$) does not apply directly.
Switch to $y$ For $x\ge0$, $y=x^{2/3}\iff x=y^{3/2}$, $0\le y\le4$. Now $\dfrac{dx}{dy}=\tfrac32y^{1/2}$ is continuous on $[0,4]$: $$L=\int_0^4\sqrt{1+\tfrac94y}\,dy=\frac{8}{27}\left(10\sqrt{10}-1\right)\approx9.0734 ,$$ the same integral as Example 1.
Explain The curves $y=x^{3/2}$ ($0\le x\le4$) and $y=x^{2/3}$ ($0\le x\le8$) are reflections of each other in the line $y=x$, and reflection does not change length. ✓
✏️ Example 5 — Numerical arc length of a sine wave
Set up the integral for the length of one arch of $y=\sin x$, $0\le x\le\pi$, and estimate it with Simpson's rule using $n=6$.
💡 Show solution▶
Set up $f'(x)=\cos x$, so $$L=\int_0^{\pi}\sqrt{1+\cos^2x}\,dx .$$ This is an elliptic integral: it has no antiderivative in terms of elementary functions, so we estimate it.
Table With $\Delta x=\tfrac{\pi}{6}$ and $g(x)=\sqrt{1+\cos^2x}$:
Check A computer with many subintervals gives $L=3.82020$, so $n=6$ is already within $0.001$. Sanity: since $1\le\sqrt{1+\cos^2x}\le\sqrt2$, the length lies between $\pi\approx3.14$ (the base) and $\pi\sqrt2\approx4.44$. ✓
Area of a Surface of Revolution
Revolve the curve $y=f(x)\ge0$ about the $x$-axis. Each segment of the inscribed polygon sweeps out a frustum — a cone with its tip cut off. If a frustum has end radii $r_1$, $r_2$ and slant height $\ell$, cutting it open and flattening it gives part of an annulus, and its lateral area is
$$A_{\text{frustum}}=\pi\,(r_1+r_2)\,\ell=2\pi\cdot\frac{r_1+r_2}{2}\cdot\ell ,$$
the circumference at the average radius times the slant height. For the $i$-th band, $r_1=f(x_{i-1})$, $r_2=f(x_i)$ and $\ell=\sqrt{1+f'(x_i^*)^2}\,\Delta x$. As $\Delta x\to0$ both radii approach $f(x_i^*)$, and the sum of band areas becomes an integral.
📐 Theorem 6.8 — Surface area of revolution
Let $f$ be smooth and nonnegative on $[a,b]$. The area of the surface obtained by revolving $y=f(x)$, $a\le x\le b$, about the $x$-axis is
$$S=\int_a^b2\pi f(x)\sqrt{1+\big(f'(x)\big)^2}\,dx .$$
If $0\le a\lt b$ and the same curve is revolved about the $y$-axis, the radius is $x$ instead of $f(x)$:
$$S=\int_a^b2\pi x\sqrt{1+\big(f'(x)\big)^2}\,dx .$$
Memory aid: $S=\displaystyle\int2\pi\,(\text{radius})\,ds$ — circumference times a little piece of arc length.
Interactive 2 — Surface Area from Frustum Bands
The curve is revolved about the $x$-axis and the surface is cut into $n$ bands. The red band is a frustum with radii $r_1,r_2$ and slant height $\ell$, so its area is $\pi(r_1+r_2)\ell$. Add all the bands and increase $n$: the total converges to $S=\int2\pi f\sqrt{1+(f')^2}\,dx$. (For the cone the bands are exact for every $n$ — why?)
6
red band: $r_1,\ r_2$—
slant height $\ell$—
band area $\pi(r_1+r_2)\ell$—
sum of $n$ bands—
exact $S$—
✏️ Example 6 — A paraboloid band, and the sphere
(a) Find the area of the surface obtained by revolving $y=\sqrt x$, $1\le x\le4$, about the $x$-axis. (b) Show that a sphere of radius $R$ has surface area $4\pi R^2$.
💡 Show solution▶
(a) Simplify $f'(x)=\dfrac{1}{2\sqrt x}$, so $1+(f')^2=1+\dfrac1{4x}=\dfrac{4x+1}{4x}$ and $$f(x)\sqrt{1+(f')^2}=\sqrt x\cdot\frac{\sqrt{4x+1}}{2\sqrt x}=\frac{\sqrt{4x+1}}{2}.$$ The awkward $\sqrt x$ cancels — this happens often in surface-area problems.
(b) Sphere Revolve $f(x)=\sqrt{R^2-x^2}$, $-R\le x\le R$. Then $f'(x)=\dfrac{-x}{\sqrt{R^2-x^2}}$ and $$f\sqrt{1+(f')^2}=\sqrt{R^2-x^2}\cdot\sqrt{\frac{R^2-x^2+x^2}{R^2-x^2}}=R .$$ So $S=\displaystyle\int_{-R}^R2\pi R\,dx=4\pi R^2$. (Strictly, $f'$ is unbounded at $x=\pm R$, so this is an improper integral; the integrand is the constant $2\pi R$, and the limit is $4\pi R^2$.)
Bonus The same calculation over $[c,c+h]\subset[-R,R]$ gives $2\pi Rh$: a slice of a sphere of thickness $h$ has the same curved area wherever it is cut — the "zone" theorem Archimedes proudly asked to have carved on his tomb.
🧭 Problem-solving strategy — arc length and surface area
Choose the variable. If $dy/dx$ blows up at an end point (vertical tangent), try writing $x=g(y)$ and integrating in $y$.
Compute $1+(\text{derivative})^2$ and simplify before taking the root. Look for a perfect square (Example 2), the identity $1+\sinh^2=\cosh^2$ (Example 3), or $1+\tan^2=\sec^2$.
For surface area, multiply by $2\pi\times$ radius first — factors often cancel (Example 6). The radius is the distance to the axis: $f(x)$ about the $x$-axis, $x$ about the $y$-axis.
If no antiderivative appears, set up the integral carefully and use Simpson's rule (Section 5.4).
Sanity-check: arc length is at least the straight-line distance between the end points.
Exercises
I
Find the exact length of each curve.
1 $y=2x+1$, $0\le x\le3$ (check with the distance formula)
5 $f'=-\tan x$, $1+\tan^2x=\sec^2x$, and $\sec x\gt0$ on $[0,\tfrac\pi4]$: $L=\displaystyle\int_0^{\pi/4}\sec x\,dx=\ln|\sec x+\tan x|\Big|_0^{\pi/4}=\ln\left(1+\sqrt2\right)\approx0.8814$.
6 As in Example 3: $L=\displaystyle\int_0^2\cosh x\,dx=\sinh2\approx3.6269$.
7 $f'=\sinh\left(\tfrac x3\right)$, so $\sqrt{1+(f')^2}=\cosh\left(\tfrac x3\right)$: $L=\Big[3\sinh\left(\tfrac x3\right)\Big]_{-3}^{3}=6\sinh1\approx7.0512$.
8 $f'=\dfrac{-2x}{1-x^2}$, $1+(f')^2=\dfrac{\left(1-x^2\right)^2+4x^2}{\left(1-x^2\right)^2}=\dfrac{\left(1+x^2\right)^2}{\left(1-x^2\right)^2}$. So the integrand is $\dfrac{1+x^2}{1-x^2}=-1+\dfrac{2}{1-x^2}=-1+\dfrac{1}{1-x}+\dfrac{1}{1+x}$ (partial fractions, Section 5.3): $L=\Big[-x+\ln\dfrac{1+x}{1-x}\Big]_0^{1/2}=\ln3-\dfrac12\approx0.5986$.
1 Find the arc-length function $s(x)$ of $y=\tfrac23x^{3/2}$ starting at $x=0$. At which point of the curve is the distance along the curve from the origin equal to $\tfrac{52}{3}$?
2 For $y=\cosh x$ with $s(x)=\sinh x$ (Example 3), find the $x$-value at which $s=1$. Give an exact answer.
3 Use Simpson's rule with $n=4$ to estimate the length of $y=x^2$, $0\le x\le1$. Compare with the exact value $\tfrac{\sqrt5}{2}+\tfrac14\ln\left(2+\sqrt5\right)$.
4 Use Simpson's rule with $n=4$ to estimate the length of $y=e^x$, $0\le x\le1$.
5 Use Simpson's rule with $n=4$ to estimate the length of $y=\dfrac1x$, $1\le x\le2$.
6Conceptual. Explain, using the triangle inequality, why every inscribed polygon is no longer than the curve — so polygon lengths are always under-estimates.
7 Without integrating, explain why $y=\ln x$, $1\le x\le e$, has exactly the same length as the curve in part 4.
💡 Show solutions▶
1 $\sqrt{1+(f')^2}=\sqrt{1+x}$, so $s(x)=\displaystyle\int_0^x\sqrt{1+t}\,dt=\frac23\left[(1+x)^{3/2}-1\right]$. Setting $s=\tfrac{52}{3}$: $(1+x)^{3/2}=27$, so $1+x=9$, $x=8$. The point is $\left(8,\tfrac23\cdot8^{3/2}\right)=\left(8,\tfrac{32\sqrt2}{3}\right)\approx(8,\,15.085)$.
2 $\sinh x=1\iff e^x-e^{-x}=2\iff e^{2x}-2e^x-1=0$, a quadratic in $e^x$: $e^x=1+\sqrt2$ (the other root is negative). So $x=\ln\left(1+\sqrt2\right)\approx0.8814$.
3 $g(x)=\sqrt{1+4x^2}$ at $0,\tfrac14,\tfrac12,\tfrac34,1$: $1,\ 1.11803,\ 1.41421,\ 1.80278,\ 2.23607$. $S_4=\tfrac{1/4}{3}\left[1+4(1.11803)+2(1.41421)+4(1.80278)+2.23607\right]\approx1.47898$. Exact $\approx1.47894$; error about $0.00003$.
4 $L=\displaystyle\int_0^1\sqrt{1+e^{2x}}\,dx$. Values at $0,\tfrac14,\dots,1$: $1.41421,\ 1.62749,\ 1.92828,\ 2.34130,\ 2.89639$. $S_4\approx\tfrac{1}{12}\left[1.41421+4(1.62749)+2(1.92828)+4(2.34130)+2.89639\right]\approx2.00353$ (true value $2.00350$).
5 $f'=-\dfrac1{x^2}$, $L=\displaystyle\int_1^2\sqrt{1+x^{-4}}\,dx$. Values at $1,1.25,1.5,1.75,2$: $1.41421,\ 1.18727,\ 1.09432,\ 1.05196,\ 1.03078$. $S_4=\tfrac{0.25}{3}\left[1.41421+4(1.18727)+2(1.09432)+4(1.05196)+1.03078\right]\approx1.13254$ (true value $1.13209$).
6 Each segment $P_{i-1}P_i$ is the shortest path between its end points, so it is no longer than the piece of curve joining them. Adding over all segments, polygon length $\le L$. Refining a partition replaces a segment by two segments with the same ends — by the triangle inequality the length can only go up. That is why Interactive 1's polygon lengths increase towards $L$.
7 $y=\ln x$ is the inverse of $y=e^x$, so its graph is the reflection of $y=e^x$ in the line $y=x$; the point $(0,1)$ maps to $(1,0)$ and $(1,e)$ to $(e,1)$. Reflection preserves length, so both lengths equal $\approx2.0035$.
III
Find the exact area of the surface obtained by revolving the curve about the stated axis.
1 $y=2x$, $0\le x\le1$, about the $x$-axis. Check with the cone formula $\pi r\ell$.
2 $y=x^3$, $0\le x\le1$, about the $x$-axis.
3 $y=\sqrt x$, $1\le x\le4$, about the $x$-axis. (Compare with Example 6.)
4 $y=\sqrt{4-x^2}$, $-1\le x\le1$, about the $x$-axis.
5 $y=\cosh x$, $-1\le x\le1$, about the $x$-axis. (Use $\cosh^2x=\tfrac12(1+\cosh2x)$.)
6 $y=x^2$, $0\le x\le1$, about the $y$-axis.
7 $y=\dfrac{x^3}{6}+\dfrac1{2x}$, $1\le x\le2$, about the $y$-axis.
8 A sphere of radius $5$ is cut by two parallel planes $3$ and $5$ units from its centre. Find the area of the curved surface between them, and compare with $2\pi Rh$.
6 Radius $x$: $S=\displaystyle\int_0^12\pi x\sqrt{1+4x^2}\,dx$. With $w=1+4x^2$: $S=\dfrac{\pi}{4}\displaystyle\int_1^5w^{1/2}dw=\dfrac\pi6\left(5\sqrt5-1\right)\approx5.3304$.
7 From Example 2, $\sqrt{1+(f')^2}=\dfrac{x^2}{2}+\dfrac{1}{2x^2}$. $S=\displaystyle\int_1^22\pi x\left(\frac{x^2}{2}+\frac1{2x^2}\right)dx=2\pi\left[\frac{x^4}{8}+\frac12\ln x\right]_1^2=2\pi\left(\frac{15}{8}+\frac{\ln2}{2}\right)=\frac{15\pi}{4}+\pi\ln2\approx13.959$.
8 Revolve $y=\sqrt{25-x^2}$, $3\le x\le5$: integrand $2\pi\cdot5$, so $S=10\pi(5-3)=20\pi\approx62.83$. And $2\pi Rh=2\pi(5)(2)=20\pi$ ✓.
IV
Applications.
1Power cable. A cable between two pylons $60$ m apart hangs in the shape $y=50\cosh\left(\tfrac{x}{50}\right)$, $-30\le x\le30$ (metres). Find the length of cable and how far it sags below the attachment points.
2Satellite dish. A dish is the surface formed by revolving $y=\tfrac{x^2}{4}$, $0\le x\le2$ (metres), about the $y$-axis. Find the area of reflective material needed.
3Cycle path. A path over gentle ramps follows $y=0.5\sin\left(\tfrac{\pi x}{10}\right)$ metres for $0\le x\le20$. Set up the integral for the path length and estimate it with Simpson's rule ($n=8$). How much longer is it than the $20$ m horizontal distance?
4Gabriel's horn. The curve $y=\tfrac1x$, $x\ge1$, is revolved about the $x$-axis. Its volume is $\pi\displaystyle\int_1^\infty x^{-2}dx=\pi$. Show that its surface area is infinite. (Compare the integrand with $\tfrac{2\pi}{x}$.)
💡 Show solutions▶
1 $f'=\sinh\left(\tfrac x{50}\right)$, $\sqrt{1+(f')^2}=\cosh\left(\tfrac x{50}\right)$: $L=\Big[50\sinh\left(\tfrac x{50}\right)\Big]_{-30}^{30}=100\sinh0.6\approx63.67$ m. Sag $=f(30)-f(0)=50(\cosh0.6-1)\approx9.27$ m.
3 $f'=0.05\pi\cos\left(\tfrac{\pi x}{10}\right)$, $L=\displaystyle\int_0^{20}\sqrt{1+0.0025\pi^2\cos^2\left(\tfrac{\pi x}{10}\right)}\,dx$. Simpson with $n=8$ ($\Delta x=2.5$) gives $L\approx20.123$ m, only about $12$ cm ($0.6\%$) more than $20$ m.
4 $S=\displaystyle\int_1^\infty\frac{2\pi}{x}\sqrt{1+\frac1{x^4}}\,dx$. Since $\sqrt{1+x^{-4}}\ge1$, the integrand is at least $\dfrac{2\pi}{x}$, and $\displaystyle\int_1^\infty\frac{2\pi}{x}dx=\lim_{t\to\infty}2\pi\ln t=\infty$. By comparison $S=\infty$: a horn you could fill with $\pi$ cubic units of paint, but never paint its surface!
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. How much farther do you really travel on a mountain road than its map distance suggests?
Why it matters. Road engineers, race organisers and fitness apps all convert between horizontal (map) distance and distance along the ground. Famous climbs — Jebel Akhdar in Oman, Alpe d'Huez in France, Baldwin Street in New Zealand — have average gradients from $8\%$ to over $30\%$.
What to do.
Model a climb with a constant gradient $m$ (e.g. $m=0.08$ for $8\%$). Show that the length along the road for map distance $D$ is $D\sqrt{1+m^2}$, and compute the percentage extra length for $m=0.08$ and $m=0.35$.
Real roads are not straight. Sample elevation profile (map distance in km, height in m): $(0,600)$, $(1,690)$, $(2,770)$, $(3,880)$, $(4,960)$, $(5,1060)$, $(6,1140)$. Estimate the road length with the polygon formula $\sum\sqrt{\Delta x^2+\Delta y^2}$ (remember to use the same units).
Fit a smooth curve (or use Simpson on $\sqrt{1+(dy/dx)^2}$ with finite-difference slopes) and compare.
Look up an elevation profile for a road near you (many free mapping tools show one) and repeat.
Tools: Calculator, spreadsheet, a free online elevation-profile tool.
Hand in: A half-page note: your extra-distance percentages, the polygon estimate, and one sentence on whether a cyclist's speedometer (which measures distance along the road) and a map would disagree noticeably.
Going further. Why does the polygon estimate always under-estimate the true length? How would you design a sampling interval so the error is below $1$ m per km?
Instructor note: For $8\%$ the extra length is only $\approx0.32\%$; even $35\%$ gives $\approx5.9\%$. The sample profile gives $\approx6024.6$ m of road for $6000$ m of map distance (about $0.4\%$ extra) — students are usually surprised how small the difference is.
🧪 Self-check quiz
0/5 correct
Q1. The length of $y=f(x)$, $a\le x\le b$, is $\displaystyle\int_a^b(\;?\;)\,dx$, where $(\;?\;)$ is
Q2. The length of the line $y=3x+2$ from $x=0$ to $x=2$ is
Q3. Which statement about $\cosh x=\tfrac12\left(e^x+e^{-x}\right)$ is true?
Q4. An inscribed polygon with $n$ segments has length $L_n$. As $n$ increases (refining the partition), $L_n$
Q5. Revolving $y=\sqrt{9-x^2}$, $-3\le x\le3$, about the $x$-axis gives a surface of area